4.5 Integration of Rational Functions by Partial Fractions
4.5
Brian E. Veitch
Integration of Rational Functions by Partial Fractions
From algebra, we learned how to find common denominators so we can do something like this,
2
3
2(x − 3)
3(x + 1)
5x − 3
+
=
+
= 2
x+1 x−3
(x + 1)(x − 3) (x + 1)(x − 3)
x − 2x − 3
So why do we need this? We need to know how to do this in the reverse order. If we’re
5x − 3
2
3
given 2
, we need to know that it can be rewritten as
+
.
x − 2x − 3
x+1 x−3
Why you ask?
Z
Let’s consider the integral
x2
5x − 3
dx.
− 2x − 3
We don’t have a method that can do this. We can’t use u-substitution, trig substitution,
integration by parts, and there are no powers of trig functions. But what if we wrote the
integral as,
Z
5x − 3
dx =
2
x − 2x − 3
Z
2
3
+
dx
x+1 x−3
Now we can evaluate this,
Z
2
3
+
dx = 2 ln |x + 1| + 3 ln |x − 3| + C
x+1 x−3
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4.5 Integration of Rational Functions by Partial Fractions
The next objective is given a rational function like
4.5.1
Brian E. Veitch
5x − 3
, how do we break it up?
x2 − 2x − 3
Method of Partial Fractions
1. Factor the denominator
x2 − 2x − 3 = (x + 1)(x − 3)
2. Next, rewrite the rational function as
x2
5x − 3
A
B
=
+
− 2x − 3
x+1 x−3
The numerators are A and B because the denominators are linear factors.
3. Clear denominators by multiplying by (x + 1)(x − 3)
5x − 3 = A(x − 3) + B(x + 1)
5x − 3 = Ax − 3A + Bx + B
5x − 3 = (A + B)x + (−3A + B)
4. Match the coefficients
5 = A+B
−3 = −3A + B
149
4.5 Integration of Rational Functions by Partial Fractions
Brian E. Veitch
5. Solve the resulting systems of equations by
(a) Using the substitution method
(b) Using the addition method
(c) Using row reducing with matrices
Solution: A = 2 and B = 3.
Therefore,
x2
5x − 3
2
3
=
+
− 2x − 3
x+1 x−3
The next methods require the degree of the numerator to be less than the degree of
the denominator. If the degree of the numerator is the same or higher, you must do long
division before proceeding to the following methods.
4.5.2
Case 1: Denominator is a product of distinct linear factors
P (x)
, where the degree of P (x) is smaller than Q(x).
Q(x)
If the degree of P (x) is greater than or equal to Q(x), you must use long division.
Suppose you have a rational function,
Find the factors of Q(x).
Z
Example 4.11. Find
x2 + 4x + 1
dx
(x − 1)(x + 1)(x + 3)
1. Set up the fraction as,
x2 + 4x + 1
A
B
C
=
+
+
(x − 1)(x + 1)(x + 3)
x−1 x+1 x+3
150
4.5 Integration of Rational Functions by Partial Fractions
Brian E. Veitch
2. Multiply through by (x − 1)(x + 1)(x + 3).
x2 + 4x + 1 = A(x + 1)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x + 1)
3. Multiply everything out, collect like terms.
x2 + 4x + 1 = Ax2 + 4Ax + 3A + Bx2 + 2Bx − 3B + Cx2 − C
x2 + 4x + 1 = (A + B + C)x2 + (4A + 2B)x + (3A − 3B − C)
4. Match coefficients
A+B+C = 1
4A + 2B = 4
3A − 3B − C = 1
(a) Add equations (1) and (3) together and you get
4A − 2B = 2
(b) Pair this with equation (2) and solve
4A − 2B = 2
4A + 2B = 4
(c) Add the equations together
8A = 6 → A =
151
3
4
4.5 Integration of Rational Functions by Partial Fractions
5. Pluggin A =
Brian E. Veitch
3
into the other equations, we get
4
1
1
3
A = ,B = ,C = −
4
2
4
6. Rewrite the integral
Z
x2 + 4x + 1
dx =
(x − 1)(x + 1)(x + 3)
Z
3/4
1/2
1/4
+
−
dx
x−1 x+1 x+3
7. Now we can integrate
Z
4.5.3
x2 + 4x + 1
3
1
1
dx = ln |x − 1| + ln |x + 1| − ln |x + 3| + C
(x − 1)(x + 1)(x + 3)
4
2
4
Case 2: Denominator is a product of repeated linear factors
If Q(x) has a factor of (x − r)n , then you have the following partial fraction breakdown,
A
B
C
+
+
+ ... +
x − r (x − r)2 (x − r)3
(x − r)n
Z
Example 4.12. Find
x4 − 2x2 + 4x + 1
dx
x3 − x2 − x + 1
1. Note the numerator has a larger degree than the numerator. You need to do long
division. After you finish, you should have
152
4.5 Integration of Rational Functions by Partial Fractions
Z
x+1+
x3
Brian E. Veitch
4x
dx
− −x+1
x2
2. We know how to integrate x + 1, so we’ll just focus on the fraction. Set up the fraction
as,
x3
A
B
C
4x
=
+
+
2
− −x+1
x − 1 (x − 1)
(x + 1)
x2
3. Multiply through by (x − 1)2 (x + 1)
4x = A(x − 1)(x + 1) + B(x + 1) + C(x − 1)2
4x = Ax2 − A + Bx + B + Cx2 − 2Cx + C
4x = (A + C)x2 + (B − 2C)x + (−A + B + C)
4. Match coefficients
A+C = 0
B − 2C = 4
−A + B + C = 0
5. Solving for A, B, C, we get A = 1, B = 2, C = −1.
6. Rewrite the integral
Z
Z
x4 − 2x2 + 4x + 1
dx =
x3 − x2 − x + 1
Z
x+1+
1
2
−1
+
+
dx
2
x − 1 (x − 1)
x+1
x4 − 2x2 + 4x + 1
1
1
dx = x2 + x + ln |x − 1| −
− ln |x + 1| + C
3
2
x −x −x+1
2
x−1
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4.5 Integration of Rational Functions by Partial Fractions
4.5.4
Brian E. Veitch
Case 3: Denominator has irreducible quadratic factors
If Q(x) in
P (x)
has irreducible factors, the partial fraction breakdown will have the term
Q(x)
Ax + B
ax2 + bx + c
where ax2 + bx + c is the irredicuble quadratic.
You can tell if a quadratic is irreducible if b2 − 4ac < 0.
Z
Example 4.13. Find
2x2 − x + 4
dx
x3 + 4x
1. Since the degree of the denominator is bigger, we can start by writing
A Bx + C
2x2 − x + 4
= + 2
2
x(x + 4)
x
x +4
2. Multiply through by x(x2 + 4)
2x2 − x + 4 = A(x2 + 4) + (Bx + C)x
2x2 − x + 4 = (A + B)x2 + Cx + 4A
3. Match the coefficients
A+B = 2
C = −1
4A = 1
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4.5 Integration of Rational Functions by Partial Fractions
Brian E. Veitch
4. Solving the system of equations we get
A = 1, B = 1, C = −1
5. Rewrite the integral
Z
2x2 − x + 4
dx =
x3 + 4x
We really can’t do much with
Z
Z
1 1x − 1
+
dx
x x2 + 4
x−1
except to break it up into two separate integrals.
x2 + 4
2x2 − x + 4
dx =
x3 + 4x
Z
1
x
1
+ 2
− 2
dx
x x +4 x +4
A quick note:
Z
1
1
−1 x
dx
=
tan
x 2 + a2
a
a
and
Z
x2
x
dx requires u-substitution
+4
6. Final Answer
Z
1 1 2
2x2 − x + 4
+ ln x + 4 − 1 tan−1 x + C
dx
=
ln
x 2
x3 + 4x
2
2
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4.5 Integration of Rational Functions by Partial Fractions
4.5.5
Brian E. Veitch
Case 4: Denonminator has repeated irreducible quadratic factors
P (x)
n
has a repeated irreducible factor (ax2 + bx + c) , the partial fraction breakQ(x)
down will have the following terms
If Q(x) in
Ax + B
Cx + D
Ex + F
+
+
+ ... +
2
2
2
2
ax + bx + c (ax + bx + c)
(ax + bx + c)3
(ax2 + bx + c)n
where ax2 + bx + c is the irredicuble quadratic.
Z
Example 4.14. Find
x2 + x + 1
dx
(x2 + 1)2
1. Write your fraction as
x2 + x + 1
Ax + B
Cx + D
=
+
(x2 + 1)2
x2 + 1
(x2 + 1)2
2. Multiply through by (x2 + 1)2
x2 + x + 1 = (Ax + B)(x2 + 1) + Cx + D
x2 + x + 1 = Ax3 + Bx2 + Ax + B + Cx + D
x2 + x + 1 = Ax3 + Bx2 + (A + C)x + (B + D)
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4.5 Integration of Rational Functions by Partial Fractions
3. Match the coefficients
A = 0
B = 1
A+C = 1
B+D = 1
4. Solving this system, we get
A = 0, B = 1, C = 1, D = 0
5. Rewrite the integral
Z
Z
6.
x2 + x + 1
dx =
(x2 + 1)2
Z
x2
1
x
+ 2
dx
+ 1 (x + 1)2
x
dx requires u-substitution.
(x2 + 1)2
Z
(x2
x
1
dx = −
+C
2
2
+ 1)
2(x + 1)
7. Final Answer:
Z
x2 + x + 1
1
dx = tan−1 (x) −
+C
2
2
2
(x + 1)
2(x + 1)
157
Brian E. Veitch